• Next 4 lectures on the BJT
• A bit to remember about transistors. Two characteristics:
1) Amplification: a small signal (I or V, terminal #1) can control a large signal (usually I, flowing between terminal #2-#3).
 Think of a tiny faucet controlling the amount of water flow through a giant hose. How many terminals? ____________.
 Why is the water faucet/hose analogy particularly useful for sub-100 nm transistors today? _________________.
2) Switching: the transistor can be turned on/off.
• Two main types of transistors studied in ECE 340:
1) Field effect transistors (FET): small input voltage (faucet) controls large output current (hose)
2) Bipolar junction transistor (BJT): small input current (faucet) controls large output current (hose)
© 2012 Eric Pop, UIUC ECE 340: Semiconductor Electronics 1

• Note, BJT = current-controlled current switch = always has some current flowing  power dissipation
• This is its Achilles’ heel compared to the MOSFET, more power dissipation  limits integration levels per chip
 BJTs have been gradually replaced with CMOS especially in highdensity logic apps over past 30 yrs.
 But, BJTs are better for analog circuit applications  higher speed and gain.
• Remember the P-N diode with lights on!
• Reverse bias (V = V p
–V n
< 0) current with lights off = small .
• When light is on, it creates________ in the middle________ region. These are swept by the E-field already present and modulate the__________.
© 2012 Eric Pop, UIUC ECE 340: Semiconductor Electronics 2

• Can we inject holes electrically rather than optically?!
 Yes, we know how to inject holes electrically. Take a p+n junction.
When forward biased, it injects holes into the n-region unilaterally:
 Now put the hole injector (forward biased p+n) and current modulator
(reverse biased np) together:
© 2012 Eric Pop, UIUC ECE 340: Semiconductor Electronics 3

• Draw like Streetman & Banerjee:
• Emitter (p+) injects holes into reverse-biased Base-Collector.
• For Collector to… collect all (most) holes we need Base to be quite_________. Mathematically, W b
<< ______.
• Properly biased, with BJT “on” the voltages are: V
EB and V
CB
< ____ (same as V
BC
> ____).
> ____
• If so, then almost all injected holes are collected: I
E
≈ I
C are the “large” currents here (water passing through the hose).
© 2012 Eric Pop, UIUC ECE 340: Semiconductor Electronics 4

• Is there a Base current I
B
? What does it consist of?
• Base current I
B
(remember base is n-type for pnp BJT):
 Some electrons____________ with injected holes.
 Some electrons injected into p+ (forward biased EB).
 Few electrons from thermal generation in reverse-biased CB junction
• Base current I
B is small, but it’s there. See figure for signs.
© 2012 Eric Pop, UIUC ECE 340: Semiconductor Electronics 5

• In a well-designed BJT, I
E
≈ I
C
≈ 100I
B
• We can write from current continuity: ______________
• There are both PNP and NPN bipolar transistors.
 Easier to study PNP because current flows in direction of holes. But
NPN is similar, replace holes w/ electrons and keep track of signs!
 Intuitively (dimensions, doping being equal), which is faster?
© 2012 Eric Pop, UIUC ECE 340: Semiconductor Electronics 6

• BJT performance parameters:
• Emitter efficiency:  
I
Ep
I
Ep

I
En
• Base transport factor:
B

I
C
/ I
Ep
• Current transfer ratio:
I
C
I
E

B
  
• Current amplification (common emitter gain):
I
C
I
B


1
 
 
© 2012 Eric Pop, UIUC ECE 340: Semiconductor Electronics 7

• Let’s begin to understand BJT physics. Think of it as two independent PN junctions:
• Band diagram in equilibrium (all I=0):
• Electric field:
• Charge density:
© 2012 Eric Pop, UIUC ECE 340: Semiconductor Electronics 8

• What did we learn so far? What is special about the Base?
• Try to make things simple(r):
 E  B current is all holes (p+ to n)
 Holes diffuse across base, ignore drift (how come?)
 Ignore any small reverse current at CB junction
 Ignore recombination (for now)
© 2012 Eric Pop, UIUC ECE 340: Semiconductor Electronics 9

• Reality check. This is what a more realistic BJT looks like:
• What kind of BJT?
• What is the cross-sectional area?
• What’s with all the extra n+ and p+ regions?
• What’s with all the SiO
2
?
© 2012 Eric Pop, UIUC ECE 340: Semiconductor Electronics 10

• Back to our simplified analysis of the P+NP (V
EB
>0, V
CB
<0):
• We know all action is minority carrier diffusion , focus there:
• In fact, all action is hole diffusion in base , get Δp there:
• Where p n
= n i
2 /N
B ECE 340: Semiconductor Electronics

• Now we can almost calculate collector (hole) current…
I
C
  qAD p dp dx x

W
B
• … because I get a nice (linear) slope from knowing W
B
<< Lp dp
 dx
• This is a narrow-base BJT, and we want it that way! Then:
I

C
• Remember, only holes are “captured” by the Collector. Why?
© 2012 Eric Pop, UIUC ECE 340: Semiconductor Electronics 12

• To get Emitter current, don’t forget there is a small electron injection component from the n-type Base.
• Draw these injected (minority) electrons in Emitter:
• And remember the expression from long-base diode:
I
En
 qA
D n
E
L
E n n i
2
N
E
 e qV
EB
/ kT 
1

• We can finally write the entire Emitter current:
I
E

I
Ep

I
En

© 2012 Eric Pop, UIUC ECE 340: Semiconductor Electronics 13

• Now we can calculate a lot of those “merit” parameters listed in the last lecture, like Emitter efficiency,
 
I I
Ep E
• Finally, what is the Base current?
I
B

I
B
(inj. to E)

I
B
(recomb. with excess holes)

I
En

Q
B
(stored)/
 p

• The current gain is
 
I
C
I
B

I
En
I
Ep

I
B , rec .

© 2012 Eric Pop, UIUC ECE 340: Semiconductor Electronics 14

• Expression for gain β can be simplified two ways:
 1) If there is no recombination in the Base, and I
B electron injection back into Emitter: is due only to

(no recomb.)

I
C

E
D L N p n E
E
I D W N
En n B B
 2) If Emitter efficiency is perfect ( γ = 1) and I
B recombination in the Base: is due only to
 
I
B
I
C
(recomb.)
2


L p
W
B


2
 Where L p
= (D p construction W
B
τ p
) 1/2 is hole diffusion length in the Base, and by
<< L p
.
 If you are not given D p how do you calculate it?
© 2012 Eric Pop, UIUC ECE 340: Semiconductor Electronics 15

• Clearly (by construction) β >> 1 so we get CURRENT GAIN!
• Why? (qualitatively). Draw energy band diagram, think about what happens if I pull a small current from Base (inject electrons).
© 2012 Eric Pop, UIUC ECE 340: Semiconductor Electronics 16

• Two main pictures for P+NP bipolar transistors:
© 2012 Eric Pop, UIUC ECE 340: Semiconductor Electronics 17

Recap our simple model for P+NP bipolar operation:
1) Current continuity: I
E
= I
B
+ I
C
2) When EB diode is forward biased (V
EB components:
> 0), Emitter current I
E has two a) Holes it injects into the Base (I
Ep
) b) Electrons the n-type Base injects into it (I
En
)
3) Base current has two components: a) Electrons it draws from terminal to supply I
En b) Electrons it draws from terminal to recombine with injected holes (Q
B
/ τ p
)
4) Collector current is made up of holes it manages to sweep (“collect”) from the ones injected by Emitter, minus those that have recombined in the Base, strictly speaking I
C
= I
Ep
– Q
B
/ τ p
• Is Collector current made up of any electrons?
• Does our current continuity assumption hold?
• When and what can we ignore to simplify our life?
© 2012 Eric Pop, UIUC ECE 340: Semiconductor Electronics 18

• One more look at the equations for P+NP:
I
Ep

D p qA
W
B
I
B , rec
I
En
 qA
D n
E
L
E n

Q
B , stored
 p
 qAW
B
2
 p
• Where, remember L p
= (D p
τ p
) 1/2 for holes in the Base.
n i
2  e qV
EB
/ kT
N
B n i
2  e qV
EB
/ kT
N
E n i
2
N
B
 e qV
EB
/ kT

1


1


1

• How do we keep things straight?
 Tip #1: Remember who is injecting what where (if it’s into the Base, then use Base parameters like D p
, L p
W
B and N
B
)
 Tip #2: If the terminal is “narrow” use its width instead of the minority carrier diffusion length there
 Tip #3: Q
B,stored width…) is just the area under the stored charge triangle in the Base (1/2 * height *
 Tip #4: On notation, if it’s unambiguous we don’t need the extra E/B/C super/subscript
(Base is the only n-type chunk here, so D p must be for… minority holes in the Base!)
 Tip #5: Remember the (e x - 1) factor is what multiplies the equilibrium minority carrier concentration (e.g. n i
2 / N
B
)
© 2012 Eric Pop, UIUC ECE 340: Semiconductor Electronics 19

• So, why do I have GAIN?! (I
C
/I
B
~ 100)
• Ex: P+NP bipolar with N
E
=10 18 cm -3 and N
B
=10 17 cm -3 . The quasi-neutral region width in the emitter is 1 μm and 0.2 μm in the base. Use μ n
μ p
=1000 cm 2 /Vs and
=300 cm 2 /Vs. The minority carrier lifetimes are 10 ns. Calculate the emitter efficiency, the base transport factor, and the current gain of this transistor biased in the forward active mode. Assume there is no recombination in the depletion regions. (Note: are the emitter and base “narrow” or “wide”?)
© 2012 Eric Pop, UIUC ECE 340: Semiconductor Electronics 20